Question: Let $g(x)=3 x^4+8x^3+4$. What is the absolute maximum value of $g$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4$ (Choice B) B $36$ (Choice C) C $116$ (Choice D) D $g$ has no maximum value
Answer: Let's first find the relative extremum points of $g$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $g$. The derivative of $g$ is $g'(x)=12x^2(x+2)$. $g'(x)=0$ for $x=0,-2$. $g'$ is defined for all real numbers. Therefore, our critical points are $x=0$ and $x=-2$. Our critical points divide the function's domain (which is all real numbers) into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $x< \llap{-}2$ $\llap{-}2<x<0$ $x>0$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $x<-2$ $x=-3$ $g'(-3)=-108<0$ $g$ is decreasing $\searrow$ $-2<x<0$ $x=-1$ $g'(-1)=12>0$ $g$ is increasing $\nearrow$ $x>0$ $x=1$ $g'(1)=36>0$ $g$ is increasing $\nearrow$ Now let's look at all the critical points: $x$ $g(x)$ Before After Verdict $-2$ $-12$ $\searrow$ $\nearrow$ Minimum $0$ $4$ $\nearrow$ $\nearrow$ Not an extremum Let's imagine ourselves walking on the graph of $g$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down until $(-2,-12)$, and then forever go up. This means that $\lim_{x\to-\infty}g(x)=\lim_{x\to +\infty}g(x)=+\infty$, which means $g$ has no maximum value. In conclusion, $g$ has no absolute maximum value.